Question

Suppose the cubic $x^3-px+q$ has $3$ distinct real roots where $p>0$ and $q>0$. Then which one of the following holds?

• A. The cubic has minima at $-\sqrt{\frac{p}{3}}$ and maxima at $\sqrt{\frac{p}{3}}$
• B. The cubic has minima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$
• C. The cubic has maxima at both $\sqrt{\frac{p}{3}}$ and $-\sqrt{\frac{p}{3}}$
• D. The cubic has minima at both $\sqrt{\frac{p}{3}}$ and maxima at $-\sqrt{\frac{p}{3}}$

Answer 1

The correct option is D

The cubic has minima at both $\sqrt{\frac{p}{3}}$ and maxima at $-\sqrt{\frac{p}{3}}$

Explanation for the correct option :

Step-1 : Finding the critical points

Let $f\left(x\right)=x^3-px+q$.

Differentiating $f\left(x\right)$ with respect to $x$, we get $f\text{'}\left(x\right)=3x^2-p$.

We know that the critical points are the roots of $f\text{'}\left(x\right)$ i.e. the solutions of $f\text{'}\left(x\right)=0$.

Now,

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ \Rightarrow 3x^2-p=0 \\ \Rightarrow x^2=\frac{p}{3} \\ \Rightarrow x=\pm \sqrt{\frac{p}{3}} \end{gathered}$$

Thus critical points of $f\left(x\right)$ are $x=\sqrt{\frac{p}{3}}$ and $x=-\sqrt{\frac{p}{3}}$.

Step-2 : Finding the point of minima and maxima

Differentiating $f\text{'}\left(x\right)$ with respect to $x$, we get $f\text{'}\text{'}\left(x\right)=6x$.

From the second derivative test, we know that $x=a$ will be a point of maxima of $f\left(x\right)$ if $f\text{'}\text{'}\left(a\right)<0$ and $x=b$ will be a point of minima of $f\left(x\right)$ if $f\text{'}\text{'}\left(b\right)>0$.

Here, as $p>0$, we have $f\text{'}\text{'}\left(\sqrt{\frac{p}{3}}\right)=6\sqrt{\frac{p}{3}}>0$. So, $x=\sqrt{\frac{p}{3}}$ is a point of minima of $f\left(x\right)$.

Again, $f\text{'}\text{'}\left(-\sqrt{\frac{p}{3}}\right)=-6\sqrt{\frac{p}{3}}<0$. So, $x=-\sqrt{\frac{p}{3}}$ is a point of maxima of $f\left(x\right)$.

Hence, option (D) is the correct answer.